# Isn't my book doing this problem about ionic product wrongly?

Problem:

After adding a $$\pu{150 mL}$$ $$\pu{6.7\times10^{-5} M}$$ $$\ce{NaOH}$$ solution to a $$\pu{100 mL}$$ $$\pu{2.5\times10^{-5} M}$$ $$\ce{FeCl2}$$ solution, green colored $$\ce{Fe(OH)2}$$ precipitate is found. What is the ionic product of $$\ce{Fe(OH)2}$$?

My book's solution:

The concentration of $$\ce{Fe^2+}$$ will be the same as the concentration of $$\ce{FeCl2}$$. So, using the formula for dilution we can get the concentration of $$\ce{Fe^2+}$$:

\begin{align} M_1V_1 &= M_2V_2 \\ 2.5\times10^5\times100 &= M_2\times250 \\ M_2 &= \frac{100\times2.5\times10^{-5}}{250} \\ &= \pu{1\times10^{-5} M} \end{align}

In the same way, the concentration of $$\ce{OH-}$$ will be the same as the concentration of $$\ce{NaOH}$$. So, using the formula for dilution we can get the concentration of $$\ce{OH-}$$:

\begin{align} M_3V_3 &= M_4V_4 \\ M_4 &= \frac{150 \times 6.7 \times 10^{-5}}{250} \\ &= \pu{4.02\times10^-5 M} \end{align}

Now, ionic product of $$\ce{Fe(OH)2}$$, $$K_\mathrm{ip}$$,

$$K_\mathrm{ip} = [\ce{Fe^2+}][\ce{OH-}]^2 =1.616\times10^{-14} \quad \text{[Ans.]}$$

My observations:

My book assumed that $$\ce{Fe(OH)2}$$ remains completely dissociated and thus the concentration of $$\ce{Fe^2+}$$ will be the same as the concentration of $$\ce{FeCl2}$$, and the concentration of $$\ce{OH-}$$ will be the same as the concentration of $$\ce{NaOH}$$. My book's answer is correct only at the very beginning of the process when any $$\ce{Fe(OH)2}$$ precipitate hasn't formed yet. However, after the precipitate has formed, isn't my book's answer wrong?

• For better site experience, you can find useful how-can-i-format-math-chemistry-expressions-here. ( Not to be applied to titles ). See also upright vs italic Nov 8, 2021 at 10:29
• The textbook uses a simplified approach with a badly worded task. It addresses the point just before precipitation start. In fact, there should be rather inequality $K_\mathrm{sp} \le \pu{1.616E−14}$, as it does not say when prcipitation started if mixing was progressive. The constant also does not count with ions trapped in precipitation. Nov 8, 2021 at 10:34
• @Poutnik Edited the question using \ce and \pu using the link you kindly provided. Moreover, if you post this comment as an answer, I'll accept it as my answer.
– user60158
Nov 8, 2021 at 12:11
• Poutnik is right. The problem should have been better written according to : When mixing small amounts of a $\ce{6.7·10^{-5} M}$ $\ce{NaOH}$ in $100$ mL $\ce{2.5·10^{-5}}$ $\ce{ FeCl2}$ solution, nothing happens. But when $150$ mL of this $\ce{NaOH}$ has been added, the first $\ce{Fe(OH)2}$ precipitate becomes visible. Calculate the ionic product of $\ce{Fe{OH}2}$ . Nov 8, 2021 at 12:38

It addresses the point just before precipitation start. In fact, there should be rather inequality $$K_\mathrm{sp} \lt \pu{1.616E−14}$$, as it does not say when prcipitation started if mixing was progressive.